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(a) We take
x
0
= 0, and solve
x
=
v
0
t
+
1
2
at
2
(Eq. 215) for the acceleration:
a
= 2(
x
–
v
0
t
)/
t
2
. Substituting
x
= 24.0 m,
v
0
= 56.0 km/h = 15.55 m/s and
t
= 2.00 s, we find
()
(
)
2
2
2 24.0m
15.55m/s
2.00s
3.56m/s ,
2.00s
a
−
==
−
or
2
  3.56 m/s
a
=
. The negative sign indicates that the acceleration is opposite to the
direction of motion of the car. The car is slowing down.
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 Spring '06
 Stoler
 Physics, Acceleration

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